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How to reverse the $n$ choose $k$ formula?

Given integers $y\geq 0$ and $z>0$, is there a good way to find an integer $x\geq y$ such that $z=\binom x y$?

I could just guess and check a few values for x because in practice its range is relatively small but this makes me feel dumb.

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binn
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  • someone made an edit that makes it seem like i want to solve for x given x ... but on the other hand maybe my original title was not more clear – binn Sep 02 '12 at 17:34
  • I've added the question to the body of the post (which is preferable to simply putting it in the title, for future reference), and rephrased it slightly for clarity. Please make certain that I've phrased it in the same way you intended it. – Cameron Buie Sep 02 '12 at 17:35
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    I'd just suggest refining guess-and-check to binary search. – Harry Altman Sep 02 '12 at 17:36
  • You might want to change the title to something like "Algorithm question involving binomial coefficients". That gives people an idea what they're looking at, and then you can give the details in the body of the post. – Cameron Buie Sep 02 '12 at 17:38
  • @cameron: thanks, your rephrasing in the question body seems accurate. – binn Sep 02 '12 at 17:39
  • @harry: the function is monotonic so i can do binary search but this still makes me feel stupid, possibly even stupider than using linear search. if there is not a slick way to get this then i will use linear search. – binn Sep 02 '12 at 17:46
  • Also, the equality gives a polynomial in $x$ of degree $y$. So there are at most $y$ such $x$'s. –  Sep 02 '12 at 17:52
  • @jennifer: i'm pretty sure there is exactly one such x with the conditions in my question, but there are possibly more if you want to have negative or fractional counts or extend factorials to complex gamma functions – binn Sep 02 '12 at 18:01
  • The sequence for varying $x$ is increasing so get an upper bound and then do a binary search. – i. m. soloveichik Sep 02 '12 at 18:06

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The first thing that comes to mind is this: $z = {x\choose y}$ has the form $x^y\over y! $ plus some lower-order terms. So calculate $x' = (y!z)^{1/y}$ and check integers that are around $x'$ to see if they work. That is a linear search, but it is a linear search that starts in almost the right place, so it is not a completely dumb linear search. If it is not smart enough, do the same thing with ${x\choose y}\approx \frac1{y!}x^{y-1}(x - y)$ instead.

I wrote a little code to do this and it seems to work reasonably well.

MJD
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  • Just to clarify, by x' = y!z^1/y I guess you mean x' = (y!z)^1/y , and by "check integers that are around x' [...] That is a linear search" you mean the linear search can be started at ceil(x') without missing the solution. – binn Sep 02 '12 at 18:48
  • Yes, just so. I have inserted the missing parentheses into the answer. By "around" I meant "I think it is a close upper bound, but I have not looked at it carefully enough to be completely sure that it is always greater than the correct answer." – MJD Sep 02 '12 at 18:56